SOLVING ALGEBRA PROBLEMS - LINKS TO THEIR STUDY, SOLVED EXAMPLES, EXERCISES ON LOGARITHMS

Please study

Logarithm Formulas before Solving Algebra Problems

if you have not already done so.

Having the knowledge of the Formulae
is a prerequisite here.

Solved Example 1 of Solving Algebra Problems :

Find the value of
(i) log₃₄₃ 49 (ii) log_0.01 (0.0001) (iii) log_(2√3) (1728)

Solution to Example 1 of Solving Algebra Problems :

(i) Let A = log₃₄₃ 49
We know 49 = 7 x 7 = 7²; 343 = 49 x 7 = 7 x 7 x 7 = 7³
∴ A = log₃₄₃ 49 = log_(7³) (7²)
We know, in log of a power to the base of another power (See Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = (2⁄3) log₇ 7
But log of any number to the same Base is one (see Formula 4).
∴ A = (2⁄3) (1) = 2⁄3. Ans.

(ii) Let A = log_0.01 (0.0001)
We know 0.0001 = 0.01 x 0.01 = (0.01)²;
∴ A = log_(0.01) (0.0001) = log_(0.01) {(0.01)²}
We know, in log of a power (see Formula 7), the exponent multiplies the log.
∴ A = 2 log_(0.01) (0.01)
But log of any number to the same Base is one (see Formula 4).
∴ A = 2 (1) = 2. Ans.

(iii) Let A = log_(2√3) (1728)
We know 2√3 = 2 x (3)^(1⁄2) = (2²)^(1⁄2) x (3)^(1⁄2)
= 4^(1⁄2) x (3)^(1⁄2) = (4 x 3)^(1⁄2) = 12^(1⁄2)
1728 = 12 x 144 = 12 x 12 x 12 = 12³
See how the base (2√3) and 1728 are made as powers of the same number 12.
∴ A = log_(2√3) (1728) = log_{12^(1⁄2)} (12³)
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = {3⁄(1⁄2)} log₁₂ 12
But log of any number to the same Base is one (see Formula 4).
∴ A = (3 x 2) (1) = 6.
Thus log_(2√3) (1728) = 6. Ans.

Solved Example 2 of Solving Algebra Problems :

If (2.3)^x = (0.23)^y = 1000, show that 1⁄x - 1⁄y = 1⁄3

Solution to Example 2 of Solving Algebra Problems :

By data (2.3)^x = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
x = log_(2.3) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄x
Reciprocal of R.H.S. = 1⁄(log_(2.3) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log₁₀₀₀ (2.3)
∴ 1⁄x = log₁₀₀₀ (2.3)

By data (0.23)^y = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
y = log_(0.23) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄y
Reciprocal of R.H.S. = 1⁄(log_(0.23) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log₁₀₀₀ (0.23)
∴ 1⁄y = log₁₀₀₀ (0.23)

To prove 1⁄x - 1⁄y = 1⁄3
L.H.S. = 1⁄x - 1⁄y = log₁₀₀₀ (2.3) - log₁₀₀₀ (0.23)
We know, log of a quotient can be written as the difference of the log of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ L.H.S. = log₁₀₀₀ {(2.3)⁄(0.23)} = log₁₀₀₀ (10)
= log_10³ (10¹)
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ L.H.S. = (1⁄3) log₁₀ 10
We know log of a quantity to the same base is 1. (See Formula 4)
∴ L.H.S. = (1⁄3) (1) = 1⁄3 = R.H.S. (Proved.)

Logarithm Tables : Characteristic and Mantissa

To find the log value of a number using Tables
we need the knowledge of Characteristic and Mantissa
which is covered in Logarithm Tables.

Exercise : Solving Algebra Problems

1. If (log₂ a)⁄4 = (log₂ b)⁄6 = (log₂ c)⁄(3p) and a³.b².c = 1,
   find the value of p
   
2. If 2 log_x a + log_ax a + 3 log_a²x a = 0,
   find the value of x.

For Answers see at the bottom of the page.

Answers to Exercise : Solving Algebra Problems

(1) -8 (2) a^-1⁄2, a^-4⁄3